# Use binomial theorem to calculate n power of a sum (MIT Single Variable Calculus)

During the first lecture of the MIT course (Single Variable Calculus - rate of Change, 46:00) professor David Jerison uses the binomial theorem to explain the following function:

$(x + \Delta x)^n = x^n + nx^{n-1} \Delta x + O((\Delta x)^2)$

Where $O((\Delta x)^2)$ is an approximation of all the remaining terms.

I understand that $(x + \Delta x)^2$ can be decomposed into $x^2 + 2x\Delta x + \Delta x^2$ but how can I understand it for n?

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By using the binomial theorem? – user2345215 Feb 27 '14 at 22:55

Hint: By the Binomial Theorem\begin{align*}(x+\Delta x)^n=&\dbinom{n}{n}x^n(\Delta x)^0+\dbinom{n}{n-1}x^{n-1}(\Delta x)^1+\dbinom{n}{n-2}x^n\cdot(\Delta x)^2+\\+&\dbinom{n}{n-3}x^n\cdot(\Delta x)^3+\ldots\end{align*} Because $\Delta x$ is a small quantity then we have that $$(\Delta x)^2>>(\Delta x)^k$$ for all $k>2$. So we say that all the other terms depend mainly on $(\Delta x)^2$ since everything else (i.e $(\Delta x)^k,$ for $k>2$) is negligible compared to it.

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This was the most complete answer, thanks a lot! – jdecuyper Feb 27 '14 at 23:13
Your welcome :)! – Jimmy R. Feb 27 '14 at 23:14

We have

$$(x+\Delta x)^n=\sum_{k=0}^n{n\choose k} x^{n-k}(\Delta x)^k ={n\choose 0}x^n+{n\choose 1}x^{n-1}\Delta x+\underbrace{\sum_{k=2}^n{n\choose k} x^{n-k}(\Delta x)^k}_{O((\Delta x)^2)}\\=x^n+nx^{n-1}\Delta x+O((\Delta x)^2)$$

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I get the 2 first terms but why is the last one $(\Delta x)^2$ and not $(\Delta x)^k$? Thank you! – jdecuyper Feb 27 '14 at 23:07
The first term of the remainder sum is for $k=2$ and it's ${n\choose 2} x^{n-2}(\Delta x)^2=O((\Delta x)^2)$ and the other terms for $k>2$ are negligible. – user63181 Feb 27 '14 at 23:13